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dV/dS*v = a

 
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Jan8-13, 03:45 AM   #1
 

dV/dS*v = a


dV/dS*v=a
now at the higest point when we throw a ball.
v=0
which implies a=0

but that is npot true....any expanations?
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Jan8-13, 03:58 AM   #2
 
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hi carhah! welcome to pf!
Quote by carhah View Post
v=0
which implies a=0
no, v= 0 does not imply dv/dt = 0

(draw a graph of v against t)
Jan8-13, 04:04 AM   #3
 
but if we put v=0 in equation-dV/dS*v=a

we get a=0?

i am in doubt...
Jan8-13, 04:06 AM   #4
 

dV/dS*v = a


Quote by tiny-tim View Post
hi carhah! welcome to pf!


no, v= 0 does not imply dv/dt = 0

(draw a graph of v against t)

but if we put v=0 in equation-dV/dS*v=a
Jan8-13, 04:26 AM   #5
 
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hi carhah!

(you don't need to send a pm … everyone gets automatic email notification of any new post in any thread they've posted in )
Quote by carhah View Post
but if we put v=0 in equation-dV/dS*v=a
but dv/ds = ∞ (draw the graph of v against s)

eg, if a is constant, and if so = vo = 0, then s = 1/2 at2 and v = at,

so v = √(2as), so dv/ds|t=0 = √(a/so)= √(a/0) = ∞
Jan8-13, 04:35 AM   #6
 
Quote by tiny-tim View Post
hi carhah!

(you don't need to send a pm … everyone gets automatic email notification of any new post in any thread they've posted in )


but dv/ds = ∞ (draw the graph of v against s)

eg, if a is constant, and if so = vo = 0, then s = 1/2 at2 and v = at,

so v = √(2as), so dv/ds|t=0 = √(a/so)= √(a/0) = ∞



thanks tim :D
BUT DOES that mean that we cannot apply that formula a=dv/ds*v in case of constant acceleration. :)
Jan8-13, 04:42 AM   #7
 
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Quote by carhah View Post
thanks tim :D
BUT DOES that mean that we cannot apply that formula a=dv/ds*v in case of constant acceleration. :)
no, you can always apply that formula …
but if it says a = ∞*0, that doesn't help very much!
Jan8-13, 05:02 AM   #8
 
can u draw graph of dv/ds ? and prove it is infinity?
Jan8-13, 05:26 AM   #9
 
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Quote by carhah View Post
can u draw graph of dv/ds ?
you can draw a graph of v against s (as i've already asked you to) …
where it's vertical, dv/ds = ∞
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